To be clear, I mean that as an actual question, not a rhetorical one. It’s a question with lots of reasonable, subjective answers. The Mets have been lucky, because **David Wright** has rediscovered his stardom, and **Kirk Nieuwenhuis** has unexpectedly settled right into the Major League lineup, and **Johan Santana**‘s left arm is not only still attached to his body seven starts into the season, but he’s arguably pitching as well as he ever has for the Mets. The Mets have also been unlucky, because they lost **Mike Pelfrey** to Tommy John surgery, and **Ike Davis** and most of the bullpen have struggled mightily, and half of their Opening Day lineup has spent time or is currently on the Disabled List.

Putting all of those details aside, though, the narrative among many about the Mets’ luck in the 2012 season has been based on the fact that they’re four games over .500, yet they’re sporting a hefty negative run differential. Specifically, they’ve surrendered 17 more runs than they’ve scored. Based on Pythagorean Expectation (the most commonly referenced way to calculate expected winning percentage) a team with 146 runs scored and 163 runs surrendered (i.e. the 2012 Mets) can expect to have a winning percentage of .445. This is calculated by taking the square of runs scored over the sum of the square of runs scored and the square of runs surrendered. A .445 winning percentage over 34 games works out to about four games UNDER .500. As we all know, in reality, the 2012 Mets are 4 games OVER .500. The problem I have with saying that the Mets are four full games better than they should be, is that it treats all runs equally. That’s fine on a long enough timeline, but on a short one, such as the first 21% of a baseball season, I think that there’s enough information ignored by the Pythagorean Expectation to make it worthwhile to explore finding a better way.

**Enter Win Probability**. For the uninitiated, Win Probability (which is freely available at FanGraphs) is the expectation that the home team will win a baseball game at any given point in that game (intuitively, one minus the home team’s win probability equals the visitors’ win probability). It is based on historical data drawn from actual Major League games. One of the benefits of having a large league that plays a ton of games every year is that there is enough data for a statistic like win probability to actually provide useful, credible information. As an example (a recent, painful example), if the home team is losing a game by two runs heading into the bottom of the ninth, they have a win expectation of 7.8%. This means that 7.8% of the time, the home team comes back to win the game (and 92.2% of the time, they lose). If however, the leadoff hitter hits a triple to start the inning, their win probability shoots up to 20.3%. A walk to the next hitter sends it up to 34.3%. And a single that plates the leadoff hitter and moves the tying run to third with still nobody out brings the home team’s win probability all the way to 64.7%, and so on and so forth. These win probabilities are calculated after every play of every game, so looking at them graphically can give you an impression of the flow of the game.

That’s where I saw an opportunity to use that information to calculate a more accurate expected record. In yesterday’s brutal 8-4 loss to the Marlins, the Mets were actually even, or in control, for most of that game. As such, from a purely “expected” vantage point, they should be credited with more than half a win for that game. However given the nature of the loss (a walkoff grand slam after the team tied up the game), Pythagorean expectation only credits the Mets with 20% of a win (4^2/(4^2+8^2)). If instead, you take the Mets’ win probability at every play, and average it over the course of the game, you’d win up with a win expectancy of 59%. In my opinion, 59% captures the story of that game (again, from a purely “expected” vantage point) a lot better than 20%.

Getting back to the original question of how lucky the Mets have been, performing this calculation for each of the Mets’ 34 games so far indicates that the Mets’ 2012 Win Probability-Based Winning Percentage is .485, which over 34 games works out to about a game under .500. The Mets have still outperformed that expectation, indicating that they’ve been lucky (or fortunate, or clutch, or however you want to label it). But compared to the team’s Pythagorean Expectation, they’ve been less lucky than it seemed.

The only other point I’d like to make about using win probability to determine expected record is that it’s purely a descriptive statistic. It isn’t useful in predicting how the Mets will perform going forward, it only fleshes out the story of how they’ve gotten to where they are. But when the question is “How lucky have the Mets been?” and the intention is to find a narrative to fit the answer, using Win Probability-Based Win Expectancy seems to me like the perfect tool. It’s more tedious to calculate than Pythagorean Expectation, but it’s also a whole lot less blunt.

*If you’re interested in Major League transactions, rules, and procedures, or if you just want to know which Mets have options left and what picks the Mets will get for Jose Reyes, be sure to check out http://tpgmets.blogspot.com and follow me on Twitter @tpgMets.*

FWIW – the Mets are third in team WPA, trailing only the Braves and Dodgers.

Well, Mets hitters rank 3rd in MLB in WPA. Unforfunately, Mets pitchers rank 23rd.

Important to highlight re: pythagorean record – in 3 games the Mets got outscored 18-9 by the Rockies, 14-6 by the Braves, and 8-1 by the Astros. That represents a -24 run differential from a small sample of lopsided losses. Two of those games ended with Chris Schwinden and Manny Acosta giving up 20 earned runs.

Sure. That’s a good example of where Pythagorean Expectation fails. With WPs, the Mets ‘ expected winning percentage for those 3 games is .177. With Pythag, it’s .137. BUT with WPs, the expectation for those 3 games is confined to those 3 games. With Pythag that effect gets spread over the rest of the season.

For instance, let’s say those 3 games, and 3 game sweep of Atlanta to start the season (1-0, 4-2, 7-5) were the Mets’ entire body of work. Pythag says the Mets expected winning % should be .262, whereas average win probabilities say it should be .445. The sub-.500 expected winning % shows how win probabilities recognize that the wins were closer games than the losses (i.e. they were closer to being losses than the losses were to being wins), but they also recognize that the runs occurred in defined games, as opposed to over the season as a whole, and as a result aren’t overly punitive.

actually, if the mets are 4 games over .500, and they are supposed to be 4 games under .500, wouldn’t that put them at 8 games better than they are supposed to be?

Each game difference is treated as 0.5

Check the current NL Standings. ATL is 22-13 and WSN is 21-13 and 0.5 games back

Exactly. By coincidence, the AL East has exactly what we’re looking for. The Yankees are 19-15 (4 games over .500), and Boston is 15-19 (4 games under .500). But in the standings, the Yankees trail Baltimore by 2.5 games, and Boston trails Baltimore by 6.5 games, putting Boston 4 games behind the Yankees.

Sidenote: The Yankees & Boston both trail Baltimore in the standings. (!!!)

i understand that when related to ANOTHER team, each win is calculated as .5, and the loss of the OTHER team calculated as the other .5, but if you are comparing just the mets wins/losses, and a team is 4 games over .500, when they are supposed to be under by 4, that is an 8 game difference.

At this point its just splitting hairs, but since I started the thread, what the heck, lemme ride it till the wheels fall off.

Lets go Mets

Fair enough. But would you say the Mets’ 2012 start was 8 games better than it would have been if they had instead been swept by Atlanta and then lost the first game to Washington?

Ultimately you’re right that we’re just talking about semantics here. Apologies if my phrasing was confusing, but I think we’re all on the same page with what I meant.